# Methods of solving parametric equations?

I have been trying to understand how to solve the exercise. I took a look at the answer and apparently you are supposed to solve $y(t) = 0$.

I do not understand this at all, I tried watching some videos on parametric equations but it just doesn‘t seem to correlate.

I mean, yes in a normal function I would understand if we were to equal the function to zero and plug the resulting $x$ into the original equation to find the y-coordinates. But why would we only search for $y(t)=0$ here and not $x(t)=0$

I would appreciate some tips on how to solve these, thanks in advance!

• You are correct in saying $y(t)=0$. Now you just need to plug in $y(t)= \sin{2t} - 2\cos{t} = 0$ and find a way to calculate the value of the expression $\cos{2t} - 3\sin{t} - 1$. This can be done in many ways, for example replacing $\sin{2t}$ by $2\sin{x}\cos{x}$. – Matti P. May 15 '18 at 5:16
• Hey Matti, thanks but well I don‘t understand why we would take y(t)=0 and not say x(t)=0. Like I said I don‘t really get it. Edited the question with further elaboration..... – user549904 May 15 '18 at 5:19
• @user549904 because the question says "the curve $K$ shares two points $A$ and $B$ with the $x$-axis." When does the curve hit the $x$-axis? Precisely when the curve's $y$ values are exactly zero, i.e. $y(t)=0$. – Pixel May 15 '18 at 5:26
• It's important to understand what are the equations of the $x$-axis and the $y$-axis. The equation of the $x$-axis is $y=0$ and the equation of the $y$-axis is $x=0$. If you just look at the picture, it should become clear. – Matti P. May 15 '18 at 5:31

You have already been told $t=\frac{\pi}{2}$ is for $A$ so you just need to evaluate $x$ and $y$ there.

$$x( \frac{\pi}{2} ) = \cos \pi - 3 \sin \frac{\pi}{2} - 1 = -1 - 3 -1 = -5\\ y( \frac{\pi}{2} ) = \sin \pi - 2 \cos \frac{\pi}{2} = 0 - 0 = 0$$

So yes $A$ does intersect the $x$-axis because $y=0$ there. The x-axis has the equation $y=0$.

What is the other time for which $y(t)=0$

$$y (t) = \sin 2 t - 2 \cos t$$

What about $\frac{3\pi}{2}$? It is similar to $\frac{\pi}{2}$ but on the other side.

$$y (\frac{3\pi}{2}) = \sin (3 \pi) - 2 \cos \frac{3\pi}{2} = 0 - 0$$

Good evaluate $x ( \frac{3\pi}{2}) = -1 - 3 (-1) - 1 = 1$

• Okay that makes sense! Finally something I can understand. Thanks Husain! – user549904 May 15 '18 at 5:21